3.187 \(\int \frac{x}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=214 \[ -\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \sqrt{a^2-b^2}}+\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \sqrt{a^2-b^2}} \]

[Out]

((-I)*x*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + (I*x*Log[1 + (b*E^(I*(c + d*
x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))]/(S
qrt[a^2 - b^2]*d^2) + PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]/(Sqrt[a^2 - b^2]*d^2)

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Rubi [A]  time = 0.400171, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3321, 2264, 2190, 2279, 2391} \[ -\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{d \sqrt{a^2-b^2}}+\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{d \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Cos[c + d*x]),x]

[Out]

((-I)*x*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + (I*x*Log[1 + (b*E^(I*(c + d*
x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))]/(S
qrt[a^2 - b^2]*d^2) + PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]/(Sqrt[a^2 - b^2]*d^2)

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x}{a+b \cos (c+d x)} \, dx &=2 \int \frac{e^{i (c+d x)} x}{b+2 a e^{i (c+d x)}+b e^{2 i (c+d x)}} \, dx\\ &=\frac{(2 b) \int \frac{e^{i (c+d x)} x}{2 a-2 \sqrt{a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}-\frac{(2 b) \int \frac{e^{i (c+d x)} x}{2 a+2 \sqrt{a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i \int \log \left (1+\frac{2 b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d}-\frac{i \int \log \left (1+\frac{2 b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2} d}\\ &=-\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{2 a-2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt{a^2-b^2} d^2}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{2 a+2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt{a^2-b^2} d^2}\\ &=-\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}+\frac{i x \log \left (1+\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}+\frac{\text{Li}_2\left (-\frac{b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d^2}\\ \end{align*}

Mathematica [B]  time = 0.810934, size = 756, normalized size = 3.53 \[ \frac{i \left (\text{Li}_2\left (\frac{\left (a-i \sqrt{b^2-a^2}\right ) \left (a+b-\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a+b+\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )\right )}\right )-\text{Li}_2\left (\frac{\left (a+i \sqrt{b^2-a^2}\right ) \left (a+b-\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a+b+\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )\right )}\right )\right )+2 (c+d x) \tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )-2 \left (\cos ^{-1}\left (-\frac{a}{b}\right )+c\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )-\log \left (\frac{(a+b) \left (-i \sqrt{b^2-a^2}-a+b\right ) \left (1+i \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac{a}{b}\right )-2 i \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )\right )-\log \left (\frac{(a+b) \left (\sqrt{b^2-a^2}+i a-i b\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right )+i\right )}{b \left (\sqrt{b^2-a^2} \tan \left (\frac{1}{2} (c+d x)\right )+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac{a}{b}\right )+2 i \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )\right )+\log \left (\frac{\sqrt{b^2-a^2} e^{-\frac{1}{2} i (c+d x)}}{\sqrt{2} \sqrt{b} \sqrt{a+b \cos (c+d x)}}\right ) \left (2 i \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )-2 i \tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )+\cos ^{-1}\left (-\frac{a}{b}\right )\right )+\log \left (\frac{\sqrt{b^2-a^2} e^{\frac{1}{2} i (c+d x)}}{\sqrt{2} \sqrt{b} \sqrt{a+b \cos (c+d x)}}\right ) \left (\cos ^{-1}\left (-\frac{a}{b}\right )+2 i \left (\tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )-\tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )\right )\right )}{d^2 \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Cos[c + d*x]),x]

[Out]

(2*(c + d*x)*ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-a^2 + b^2]] - 2*(c + ArcCos[-(a/b)])*ArcTanh[((-a + b)*T
an[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + (ArcCos[-(a/b)] - (2*I)*ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-a^2 + b^
2]] + (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])*Log[Sqrt[-a^2 + b^2]/(Sqrt[2]*Sqrt[b]*E^((I
/2)*(c + d*x))*Sqrt[a + b*Cos[c + d*x]])] + (ArcCos[-(a/b)] + (2*I)*(ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-
a^2 + b^2]] - ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]))*Log[(Sqrt[-a^2 + b^2]*E^((I/2)*(c + d*x)
))/(Sqrt[2]*Sqrt[b]*Sqrt[a + b*Cos[c + d*x]])] - (ArcCos[-(a/b)] - (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/S
qrt[-a^2 + b^2]])*Log[((a + b)*(-a + b - I*Sqrt[-a^2 + b^2])*(1 + I*Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a^2 +
b^2]*Tan[(c + d*x)/2]))] - (ArcCos[-(a/b)] + (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])*Log
[((a + b)*(I*a - I*b + Sqrt[-a^2 + b^2])*(I + Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]
))] + I*(PolyLog[2, ((a - I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a
^2 + b^2]*Tan[(c + d*x)/2]))] - PolyLog[2, ((a + I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2
]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]))]))/(Sqrt[-a^2 + b^2]*d^2)

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Maple [B]  time = 0.418, size = 414, normalized size = 1.9 \begin{align*}{\frac{ix}{d}\ln \left ({ \left ( b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}-{\frac{ix}{d}\ln \left ({ \left ( -b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+{\frac{ic}{{d}^{2}}\ln \left ({ \left ( b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}-{\frac{ic}{{d}^{2}}\ln \left ({ \left ( -b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}-{\frac{1}{{d}^{2}}{\it dilog} \left ({ \left ( -b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+{\frac{1}{{d}^{2}}{\it dilog} \left ({ \left ( b{{\rm e}^{i \left ( dx+c \right ) }}+\sqrt{{a}^{2}-{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}-{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+{\frac{2\,ic}{{d}^{2}}\arctan \left ({\frac{2\,b{{\rm e}^{i \left ( dx+c \right ) }}+2\,a}{2}{\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*cos(d*x+c)),x)

[Out]

I/d/(a^2-b^2)^(1/2)*ln((b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*x-I/d/(a^2-b^2)^(1/2)*ln((-b*
exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x+I/d^2/(a^2-b^2)^(1/2)*ln((b*exp(I*(d*x+c))+(a^2-b^2)
^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*c-I/d^2/(a^2-b^2)^(1/2)*ln((-b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^
(1/2)))*c-1/d^2/(a^2-b^2)^(1/2)*dilog((-b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))+1/d^2/(a^2-b
^2)^(1/2)*dilog((b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))+2*I/d^2*c/(-a^2+b^2)^(1/2)*arctan(1/
2*(2*b*exp(I*(d*x+c))+2*a)/(-a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.37501, size = 2338, normalized size = 10.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*
a) - 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a
) + 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a
) - 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a
) + 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(
d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) + 2*I*a
*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*b*sqrt((a^2 - b^
2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 -
b^2)/b^2) + 2*b)/b + 1) - 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) - 2*(b*c
os(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(I*b*d*x + I*b*c)*sqrt((a^2 - b^2)/b^2
)*log(1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2)
+ 2*b)/b) + 2*(-I*b*d*x - I*b*c)*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(d*x + c) + 2*I*a*sin(d*x + c) - 2*(b*
cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 2*(-I*b*d*x - I*b*c)*sqrt((a^2 - b^2)/b^2)*
log(1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) +
2*b)/b) + 2*(I*b*d*x + I*b*c)*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(d*x + c) - 2*I*a*sin(d*x + c) - 2*(b*cos
(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + 2*b)/b))/((a^2 - b^2)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x)

[Out]

Integral(x/(a + b*cos(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate(x/(b*cos(d*x + c) + a), x)